Question

求使 [x^2-2x+1分之x^2+x+1-(x-1)^3分之x^3+1]·(x^2-2x+1)的值为整数的x的所有整数

Answer 1

[x^2-2x+1分之x^2+x+1-(x-1)^3分之x^3+1]·(x^2-2x+1)

=x²+x+1-(x³+1)/(x-1)

=[(x³-1)-(x³+1)]/(x-1)

=-2/(x-1)

∴整数x的值是

x-1=-2 x-1=-1 x-1=1 x-1=2

即x=-1 x=0 x=2 x=3